Question

Use the Laplace transform to solve the initial value problem. $$y^{\prime \prime}+4 y=\sin 2 t, \quad y(0)=1, \quad y^{\prime}(0)=0$$

Short answer

Question: Find the solution to the initial value problem \(y^{\prime\prime} + 4y = \sin 2t, y(0)=1, y'(0)=0\). Answer: The solution to the initial value problem is \(y(t) = \cos 2t + \frac{1}{4}(t\sin 2t)\).

Step-by-step solution

Apply Laplace Transform

Let's take Laplace transform of both sides of the given differential equation: $$\mathcal{L}\{y^{\prime\prime}+4y\} = \mathcal{L}\{\sin 2t\}$$ Using linearity property and initial conditions, we get: $$\mathcal{L}\{y^{\prime\prime}\} + 4\mathcal{L}\{y\} = s^2Y(s) - sy(0) - y'(0) + 4Y(s) = s^2Y(s) - s + 4Y(s)$$

Solve for Laplace Transform Y(s)

Now, we can solve for Y(s): $$Y(s)(s^2 + 4) = s + \frac{2}{s^2+4}$$ Now, we need to isolate Y(s) by dividing by \((s^2+4)\): $$Y(s) = \frac{s}{s^2+4} + \frac{2}{(s^2+4)^2}$$

Inverse Laplace Transform

To find the solution function y(t), we need to take the inverse Laplace transform: $$\mathcal{L}^{-1}\{Y(s)\} = \mathcal{L}^{-1}\left\{\frac{s}{s^2+4} + \frac{2}{(s^2+4)^2}\right\}$$ The inverse Laplace transform is given by: $$y(t) = \mathcal{L}^{-1}\left\{\frac{s}{s^2+4}\right\} + \mathcal{L}^{-1}\left\{\frac{2}{(s^2+4)^2}\right\}$$ We recognize the first term as the Laplace transform of \(\cos 2t\) and the second term can be evaluated using the convolution theorem, but we recognize it as the Laplace transform of \(\frac{1}{4}(t\sin 2t)\). Therefore, $$y(t) = \cos 2t + \frac{1}{4}(t\sin 2t)$$ This is the solution to the given initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They describe various phenomena in fields such as physics, engineering, and economics. The differential equation in our exercise is a second-order linear ordinary differential equation with a sine function as a non-homogeneous term.

The problem requires us to find a function, usually denoted as y(t), which satisfies the relationship given in the equation when subjected to a set of initial conditions. Here, the initial conditions are the values of the function and its first derivative at time t = 0. In this case, those conditions are: \( y(0) = 1 \) and \( y'(0) = 0 \). Usually, the process of finding the solution involves several steps, starting with identifying the type of differential equation, finding the complementary function (for the homogeneous part), and finding a particular solution (for the non-homogeneous part).

However, our exercise makes use of the Laplace transform, a powerful technique that transforms the differential equation into an algebraic equation, which is often easier to solve.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical process used to revert the Laplace transform and obtain the original function of time from its transformed counterpart, a function of complex frequency s, denoted as Y(s). The inverse transform is denoted by \( \mathcal{L}^{-1} \).

In the context of our problem, after taking the Laplace transform of both sides of the given differential equation and solving for Y(s), we are left with an expression in the s-domain. To find the time-domain solution y(t), we need to apply the inverse Laplace transform.

The inverse process involves finding a known transform pair that matches the terms in Y(s) or decomposing Y(s) into simpler fractions that can be readily inverted. Matching transform pairs often come from a table of Laplace transforms, which is a critical tool for this process. In the exercise, one term matches the transform of \( \cos(2t) \) while another term is manipulated to match the transform of \( \frac{1}{4}(t \sin(2t)) \), based on the properties of the Laplace transform.

Linearity Property of Laplace Transform

The linearity property of the Laplace transform is fundamental to solving differential equations because it allows us to handle the transform of a sum of functions as the sum of their individual transforms. Formally, if \( f(t) \) and \( g(t) \) are functions of time t, and a and b are constants, then the Laplace transform of their linear combination is given by:

\( \mathcal{L}\{af(t) + bg(t)\} = a\mathcal{L}\{f(t)\} + b\mathcal{L}\{g(t)\} \).

This property was used in the solution when the Laplace transform was applied to both terms on the left side of the differential equation separately. By exploiting this property, we can work with each term individually, which simplifies the algebra involved in solving for Y(s). It also makes it possible to handle initial conditions directly in the transformed equation, as shown in the exercise solution.

Initial Conditions

Initial conditions in a differential equation problem are the values describing the state of the system at the beginning of the observation, typically at time \( t = 0 \). The initial conditions for our exercise are \( y(0) = 1 \) and \( y'(0) = 0 \). These conditions are crucial when solving differential equations, as they allow us to find a unique solution that fits the specific scenario of the problem.

Differentially, when applying the Laplace transform to solve a differential equation, initial conditions are incorporated into the transformed equation. This is seen where the terms involving \( y(0) \) and \( y'(0) \) come into play when the Laplace transform of the derivatives is taken. Whenever the Laplace transform of a derivative is computed, we subtract the initial values of the function and its derivatives, multiplied by powers of s. These terms ensure that the transformed equation, Y(s), retains the necessary information to accurately reflect the system's initial state when computing the inverse Laplace transform to obtain the final solution.